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Probabilities of duplicates 
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#1
Mar1509, 03:39 PM

P: 2,500

What is the probability of getting a duplicate of a digit in a randomly generated sequence of four digits?
I would approach this intuitively as follows. (1.0)(0.1)(0.9)(0.9)=0.081 " (0.1)(0.1)(0.9)=0.009 " (0.1)(0.1)(0.1)=0.001 Sum =0.091 There must be a more efficient and elegant way of doing this but I've not been able to find one. Also, is there a more general formula for a set of n distinct objects each with a probability 1/n of being selected? 


#2
Mar1509, 03:53 PM

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Hi SW VandeCarr!



#3
Mar1509, 04:04 PM

P: 2,500




#4
Mar1509, 04:13 PM

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P: 26,148

Probabilities of duplicates
start again 


#5
Mar1509, 04:17 PM

P: 2,500




#6
Mar1509, 04:25 PM

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and why is there only one variable (n)? 


#7
Mar1509, 04:54 PM

P: 2,500

You're correct, there is more than one variable. From a set of N distict objects, a random sequence of n objects is generated. The probability of not getting a duplicate in the sequence of n objects is ((N1)/N)^n. Now, let r equal all replications (duplicates, three of a kind, four of a kind, etc) each considered to contain just one duplicate, what is the most efficient way to calculate the probability of a duplicate so defined? 


#8
Mar1509, 05:02 PM

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#9
Mar1509, 05:47 PM

P: 813

Well, if the number of digits is greater then the base then the probability is equal to one.



#10
Mar1509, 06:09 PM

P: 2,500

My reasoning regarding the probability of getting only a duplicate in four randomly generated digits is: The first selection can be any digit and therefore has a probability of one. The second selection has a probability of 0.1 of being the same digit. The third and fourth selection have a probability 0.9 of not being the same digit. Since multiplication is commutative, the actual order doesn't matter. If we consider the probability of getting at least a duplicate, then we need to add the probabilities of getting three or four of a kind. I don't think the answer is as simple as 1(0.9)^3. 


#11
Mar1509, 06:40 PM

P: 813

Probability of a repeat 1/B where B is the base There are now two digest taken. So when we add another diget: 1/(B)+(1/(B1))*(11/B) As there is 1/B chance of the third digest matching the fist digit, but if the third digit doesn't match the first that leaves (B1) numbers in the diget left. So the probability of it matching the second digit is 1/(B1) multiplied by the probability that it didn't match the first digit. Let P(N) be the probability of a match after N digest then [tex]P(N)=P(N1)+(1P(N1))*{1 \over BN+2}[/tex] and [tex]P(2)=1/B[/tex] 


#12
Mar1509, 09:32 PM

P: 2,500

Actually the argument is counterintuitive if one is thinking in terms of cumulative probability. For cumulative probability 1(0.9)^3 is correct. That is, once the marker digit is selected, the cumulative probability that a duplicate will be selected in the next three selections is about 0.271. However, cumulative probability is not the same as sequential probability. The probability that the marker will be matched in any particular one of the next three independent selections is still 0.1 in frequentist theory.
The idea of cumulative probability fits with experience, but an it's entirely different concept than the probability of a single independent event. I don't know the answer to this problem, but it is a problem. Randomized scientific trials use threshold probabilities to determine statistical significance but rarely adjust the significance level for the number of similar trials conducted. If you believe in cumulative probability, scientific trials should adjust these levels or be liable for false positive results. 


#13
Mar1509, 09:59 PM

P: 813

Which part of my last post do you disagree with and how does this apply to randomized scientific trials. Is there an area of research were these statistics are particularly important? 1(0.9)^3 is probably an okay approximation if the base is large relative to the number of digits. Since the base is 10 and the number of digest is four, it might be an okay approximation for this problem.
Your formula clearly fails when the number of digits N=10 since it gives a finite probability of no repeats which is impossible with 10 digits. 


#14
Mar1609, 04:10 AM

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i assumed from your original post that it was the latter 


#15
Mar1609, 10:18 AM

P: 2,500

For sequential probability, the question is: given an existing digit sequence of n digits with no duplicates, what is the probability that the next digit selected will be a duplicate of one of the previously generated digits? For a generated sequence that already has just one duplicate, the probability that an expanded sequence (by adding more randomly generated digits) will have just one duplicate decreases. 


#16
Mar1609, 11:55 AM

P: 813




#17
Mar1609, 03:43 PM

P: 2,500

The title is "probabilities of duplicates" and I'm not giving a tutorial. I'm asking questions because I don't know the answers. 


#18
Mar2409, 03:25 PM

P: 2,500

P(k(i)=(k(j)) = (n1)/B where n is the number of trials and k is the kth trial within n and (n1) less than or equal to B. ( For for the first trial P=0 and for the eleventh trial P=1 as it should.. For at most one duplicate in n trials, I suggest my first post is correct. Any comments tiny tim or others? 


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